What is the tangent of f(x)=(6x-8)^{1/2},\at x=4 ?

The tangent of f(x)=(6x-8)^{1/2},\at x=4 is y= 3/4 x+1

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tangent of f(x)=(6x-8)^{1/2},\at x=4

tangent of

f (x) = (6 x - 8)^{\frac{1}{2}}, at x = 4

y = \frac{3}{4} x + 1

Solution steps

Find the tangent point:

(4, 4)

Find the slope of

f (x) = (6 x - 8)^{\frac{1}{2}} : \frac{df ( x )}{d x} = \frac{3}{( 6 x - 8 ) ^{\frac{1}{2}}}

Compute the slope of

f (x) = (6 x - 8)^{\frac{1}{2}}

x = 4 : m = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(4, 4) : y = \frac{3}{4} x + 1

y = \frac{3}{4} x + 1

\int_{0}^{π} x^{2} cos (n x) d x

\int \frac{1}{x ^{2} x ^{2} - 7} d x

\frac{d}{d x} (x^{2})

\int_{0}^{1} \frac{2}{( x + 1 ) ( x ^{2} + 1 )} d x

\frac{d}{d x} (x^{3} + 3 x^{2} - 24 x - 80)

What is the tangent of f(x)=(6x-8)^{1/2},\at x=4 ?
The tangent of f(x)=(6x-8)^{1/2},\at x=4 is y= 3/4 x+1